"Vectors" do not have a point of intersection since they are "movable". What you want is the intersection point of two lines, one of which contains the point (0, 0, 0)has direction vector <-.41, .28, -.08>. That first line can be written in parametric equations as x= -.41t, y= .28t, z= -.08t.

Now, you are told that the second line contains the point (-2.70, -.45, -.21)
and makes an angle of 38.3 degrees with the first line. Now the problem is that in three dimensions there exist an infinite number of lines that make an angle of 38.3 degrees with that first line- Take any cone at an angle of 38.3 degree to that line at any point on the line.

I think what you can do is this: let (x0, y0, z0) be a point on the first line. A vector from that line to (-2.70, -.45, -.21) is of the form <-2.7- x0, -.45- y0, -.21- z0>.

Since the dot product of two vectors, u and v, is given by [itex]u\cdot v= |u||v|cos(\theta)[/itex] we have
[tex](-2.7- x0)(-2.7)+ (-.45- y0)(-.45)+ (-.21- z0)(-.21)[/tex][tex]= \sqrt{(-2.7- x0)^2+ (-.45- y0)^2+ (-.21- z0)^2}\sqrt{(2.7)^2+ (.45)^2+ (.21)^2}[/tex]

Replace x0, y0, and z0 with x0= -.41t, y0= .28t, z0= -.08t and that becomes an equation in the single variable t. Solve for t and use that t to find x, y, and z.